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September 30, 2014

September 30, 2014

Posted by **Lisa** on Monday, July 22, 2013 at 3:55pm.

- Trig -
**Steve**, Monday, July 22, 2013 at 4:52pm17sec^2θ - 15secθ tanθ - 15 = 0

15secθ tanθ = 17sec^2θ - 15

225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225

225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225

64sec^4θ - 285sec^2θ + 225 = 0

That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.

However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.

In radians, I get .159,1.000,2.141,2.982

- Trig -
**Reiny**, Monday, July 22, 2013 at 5:01pm17sec2 θ − 15tanθsecθ − 15 = 0

17/cos^2 Ø - 15(sinØ/cosØ)(1/cosØ) - 15 = 0

times cos^2 Ø

17 - 15sinØ - 15cos^2 Ø = 0

17 - 15sinØ - 15(1 - sin^2 Ø) = 0

15sin^2 Ø - 15sinØ + 2 = 0

sinØ = (15 ± √105)/30

sinØ = .158435 or sinØ = .841565

Ø = 9.1° or 170.9° or Ø = 57.3° or 122.7°

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